Ta có: \(p+e+n=52\)
\(\Leftrightarrow n+12+n=52\)
\(\Leftrightarrow2n+12=52\)
\(\Leftrightarrow2n=40\)
\(\Leftrightarrow n=20\)
\(\Rightarrow p+e=20+12=32\)
Mà \(p=e\Rightarrow p=e=\dfrac{1}{2}\times32=16\)
Vậy nguyên tử X có: \(p=e=16\left(hạt\right);n=20\left(hạt\right)\)